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Chemical Kinetics and Nuclear Chemistry question

2019 · 12 Apr · Shift 2 · Q11
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Chemical Kinetics and Nuclear Chemistry question

2019 · 12 Apr · Shift 2 · Q11

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
NO2NO_2NO2​ required for a reaction is produced by the decomposition of N2O5N_2O_5N2​O5​ in CCl4CCl_4CCl4​ as per the equation, 2N2O5N_2O_5N2​O5​(g) →\to→ 4NO2NO_2NO2​(g) + O2O_2O2​(g). The initial concentration of N2O5N_2O_5N2​O5​ is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2NO_2NO2​ is :
  1. A
    2.083 × 10–3 mol L–1 min–1
  2. B
    8.333 × 10–3 mol L–1 min–1
  3. C
    4.167 × 10–3 mol L–1 min–1
  4. D
    1.667 × 10–2 mol L–1 min–1
View written solutionFree

Correct answer: D

  1. Given reaction

2N2O5(g)→4NO2(g)+O2(g)2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)2N2​O5​(g)→4NO2​(g)+O2​(g)

  1. Change in concentration of N2O5N_2O_5N2​O5​

Initial concentration: [N2O5]0=3.00 mol L−1[N_2O_5]_0 = 3.00\ \text{mol L}^{-1}[N2​O5​]0​=3.00 mol L−1

Concentration after 30 min: [N2O5]30=2.75 mol L−1[N_2O_5]_{30} = 2.75\ \text{mol L}^{-1}[N2​O5​]30​=2.75 mol L−1

So, decrease in concentration in 30 min is Δ[N2O5]=3.00−2.75=0.25 mol L−1\Delta [N_2O_5] = 3.00 - 2.75 = 0.25\ \text{mol L}^{-1}Δ[N2​O5​]=3.00−2.75=0.25 mol L−1

Thus, average rate of disappearance of N2O5N_2O_5N2​O5​ is −Δ[N2O5]Δt=0.2530=8.333×10−3 mol L−1 min−1-\frac{\Delta [N_2O_5]}{\Delta t} = \frac{0.25}{30} = 8.333 \times 10^{-3}\ \text{mol L}^{-1}\text{ min}^{-1}−ΔtΔ[N2​O5​]​=300.25​=8.333×10−3 mol L−1 min−1

  1. Use stoichiometric relation

From 2N2O5→4NO22N_2O_5 \rightarrow 4NO_22N2​O5​→4NO2​

for every 2 moles of N2O5N_2O_5N2​O5​ consumed, 4 moles of NO2NO_2NO2​ are formed. So, Rate of formation of NO2=42×Rate of disappearance of N2O5\text{Rate of formation of } NO_2 = \frac{4}{2} \times \text{Rate of disappearance of } N_2O_5Rate of formation of NO2​=24​×Rate of disappearance of N2​O5​

=2×8.333×10−3= 2 \times 8.333 \times 10^{-3}=2×8.333×10−3

=1.667×10−2 mol L−1 min−1= 1.667 \times 10^{-2}\ \text{mol L}^{-1}\text{ min}^{-1}=1.667×10−2 mol L−1 min−1

  1. Match with options

1.667×10−2 mol L−1 min−11.667 \times 10^{-2}\ \text{mol L}^{-1}\text{ min}^{-1}1.667×10−2 mol L−1 min−1 corresponds to Option D.

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